Vidaara.orgClass 11 · Chemistry
CodeVID-C11-04-T3-01
Molecular Orbital Theory & Hydrogen Bonding - Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end - for full solutions, raise your doubts with your teacher.
Section A - Multiple Choice Questions
5 × 1 = 5 marks
1.
The formula for bond order in MO theory is:
- A.$N_b + N_a$
- B.$\frac{N_b - N_a}{2}$
- C.$\frac{N_a - N_b}{2}$
- D.$N_b \times N_a$
2.
Bond order of He2 is:
- A.1
- B.2
- C.0
- D.0.5
3.
A molecule with unpaired electrons is:
- A.diamagnetic
- B.paramagnetic
- C.non-magnetic
- D.ionic
4.
Hydrogen bonding occurs in:
- A.CH4
- B.H2O
- C.CO2
- D.Cl2
5.
Intramolecular hydrogen bonding, compared with intermolecular, generally:
- A.raises the boiling point
- B.lowers the boiling point
- C.has no effect
- D.makes the solid ionic
Section B - Short Answer (2 marks)
3 × 2 = 6 marks
6.
Distinguish between bonding and antibonding molecular orbitals.
7.
Calculate the bond order of N2 (Nb = 10, Na = 4).
8.
Why is ice less dense than liquid water?
Section C - Short Answer (3 marks)
2 × 3 = 6 marks
9.
Using MO theory, explain why O2 is paramagnetic while N2 is diamagnetic.
10.
Explain why HF has a higher boiling point than HCl although HCl has a larger molar mass.
Section D - Long Answer (5 marks)
1 × 5 = 5 marks
11.
Draw the molecular orbital energy diagram of O2, calculate its bond order and explain its magnetic property; then define hydrogen bonding and give two physical consequences.
Answer Key
Section A - Multiple Choice Questions
- (B) $\frac{N_b - N_a}{2}$
- (C) 0
- (B) paramagnetic
- (B) H2O
- (B) lowers the boiling point
Section B - Short Answer (2 marks)
- Bonding MOs are lower in energy with high electron density between the nuclei; antibonding MOs are higher in energy with a node between the nuclei.
- Bond order = (10 - 4)/2 = 3.
- In ice, hydrogen bonds hold water molecules in an open cage-like structure with large empty spaces, lowering the density below that of liquid water.
Section C - Short Answer (3 marks)
- In O2 the last two electrons occupy two degenerate pi* orbitals singly (Hund's rule), giving two unpaired electrons (paramagnetic). In N2 all electrons are paired, so it is diamagnetic.
- Fluorine is small and very electronegative, so HF forms strong intermolecular hydrogen bonds that need extra energy to break, raising its boiling point above HCl, which has only weaker dipole-dipole forces.
Section D - Long Answer (5 marks)
- O2 (16 e): sigma2s, sigma*2s, sigma2p, two pi2p, two pi*2p (singly filled). Bond order = (10-6)/2 = 2; two unpaired pi* electrons make it paramagnetic. Hydrogen bonding is the attraction between an H atom bonded to F/O/N and a lone pair on a neighbouring electronegative atom; consequences: high boiling point of water and the low density of ice.
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